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The Matrix Exponential via Eigenvalues and Eigenvectors

Module by: Steven J. Cox. E-mail the author

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Summary: This module introduces how to compute the matrix exponential using eigenvalues and eigenvectors.

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In this module we exploit the fact that the matrix exponential of a diagonal matrix is the diagonal matrix of element exponentials. In order to exploit it we need to recall that all matrices are almost diagonalizable. Let us begin with the clean case: if AA is nn-by-nn and has nn distinct eigenvalues, λjλj, and therefore nn linear eigenvectors, sjsj, then we note that ∀j,j∈1…n:Asj=λjsj j j 1 … n A sj λj sj may be written

A=SΛS-1 A S Λ S (1)
where S=s1s2…sn S s1 s2 … sn is the full matrix of eigenvectors and Λ=diagλ1λ2…λn Λ diag λ1 λ2 … λn is the diagonal matrix of eigenvalues. One cool reason for writing AA as in Equation 1 is that A2=SΛS-1SΛS-1=SΛ2S-1 A 2 S Λ S S Λ S S Λ 2 S and, more generally Ak=SΛkS-1 A k S Λ k S If we now plug this into the definition in The Matrix Exponential as a Sum of Powers, we find ⅇAt=SⅇΛtS-1 A t S Λ t S where ⅇΛt Λ t is simply diagⅇλ1tⅇλ2t…ⅇλnt diag λ1 t λ2 t … λn t Let us exercise this on our standard suite of examples.

Example 1

If A=1002 A 1 0 0 2 then S=IΛ=A S I Λ A and so ⅇAt=ⅇΛt A t Λ t . This was too easy!

Example 2

As a second example let us suppose A=01-10 A 0 1 -1 0 and compute, in matlab, 


	
	>> [S, Lam] = eig(A)

	   S = 0.7071             0.7071
	            0 + 0.7071i        0 - 0.7071i


	   Lam = 0 + 1.0000i     0
	         0               0 - 1.0000i


	>> Si = inv(S)

	   Si = 0.7071     0 - 0.7071i
	        0.7071     0 + 0.7071i


	>> simple(S*diag(exp(diag(Lam)*t))*Si)

	   ans = [ cos(t),   sin(t)]
	         [-sin(t),   cos(t)]

Example 3

If A=0100 A 0 1 0 0 then matlab delivers 


	
	>> [S, Lam] = eig(A)

	   S = 1.0000   -1.0000
	       0         0.0000

	   Lam = 0    0
	         0    0

So zero is a double eigenvalue with but one eigenvector. Hence SS is not invertible and we can not invoke (Equation 1). The generalization of (Equation 1) is often called the Jordan Canonical Form or the Spectral Representation. The latter reads A=∑j=1hλjPj+Dj A j 1 h λj Pj Dj where the λjλj are the distinct eigenvalues of AA while, in terms of the resolvent Rz=zI−A-1 R z z I A , Pj=12πⅈ∫CjRzdz Pj 1 2 z Cj R z is the associated eigen-projection and Dj=12πⅈ∫CjRzz−λjdz Dj 1 2 z Cj R z z λj is the associated eigen-nilpotent. In each case, CjCj is a small circle enclosing only λjλj.

Conversely we express the resolvent Rz=∑j=1h1z−λjPj+∑k=1mj−11z−λjk+1 D j k R z j 1 h 1 z λj Pj k 1 mj 1 1 z λj k 1 D j k where mj=dimℛPj mj dim ℛ Pj with this preparation we recall Cauchy's integral formula for a smooth function ff fa=12πⅈ∫Cafzz−adz f a 1 2 z C a f z z a where Ca C a is a curve enclosing the point aa. The natural matrix analog is fA=-12πⅈ∫CrfzRzdz f A -1 2 z C r f z R z where Cr C r encloses ALL of the eigenvalues of AA. For fz=ⅇzt f z z t we find

ⅇAt=∑j=1hⅇλjtPj+∑k=1mj−1tkk! D j k A t j 1 h λj t Pj k 1 mj 1 t k k D j k (2)
with regard to our example we find, h=1 h 1 , λ 1 =0 λ 1 0 , P 1 =I P 1 I , m 1 =2 m 1 2 , D 1 =A D 1 A so ⅇAt=I+tA A t I t A Let us consider a slightly bigger example, if A=110010002 A 1 1 0 0 1 0 0 0 2 then 


	
	>> R = inv(s*eye(3)-A)

	   R = [ 1/(s-1),   1/(s-1)^2,         0]
	       [       0,     1/(s-1),         0]
	       [       0,           0,   1/(s-2)]

and so λ 1 =1 λ 1 1 and λ 2 =2 λ 2 2 while P1=100010000 P1 1 0 0 0 1 0 0 0 0 and so m 1 =2 m 1 2 D1=010000000 D1 0 1 0 0 0 0 0 0 0 and P2=000000001 P2 0 0 0 0 0 0 0 0 1 and m 2 =1 m 2 1 and D 2 =0 D 2 0 . Hence ⅇAt=ⅇtP1+tD1+ⅇ2tP2 A t t P1 t D1 2 t P2 ⅇttⅇt00ⅇt000ⅇ2t t t t 0 0 t 0 0 0 2 t

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